Topics: IF If statements Else clauses

IF Statement For the conditional expression, evaluating to True or False, the simple IF statement is if : x = 7 if x > 10: print (x) x = x + 1 x = True if x: print (x)

Conditionals It is also possible to specify what to do if the condition is False. Contrast these two examples. x = 7 if x > 10: print (x) x = x + 1 x = 7 if x > 10: print (x) else: x = x + 1

if x > 10 print (x) x = x + 1 True False x = 7 if x > 10: print (x) x = x + 1

if x > 10 print (x) TrueFalse x = 7 if x > 10: print (x) else: x = x + 1

Conditions At the heart of a conditional is a condition evaluation The code after the if but before the : should evaluate to either True or False These are special keywords in Python and are case sensitive

Beyond True and False Python provides True and False, even for general values Numbers:zero ≡ False, not zero ≡ True Lists: [ ] ≡ False, any other list ≡ True Strings:“” ≡ False, any other string ≡ True Experiment with it! x = 0 if 1: print (x) x = 0 if 0: print (x)

Do these programs print the same things? # 1 x = 7 if x > 10: print (x) x = x + 1 print (x) #2 x = 7 if x > 10: print (x) else: x = x + 1 print (x) (a)Yes (b)No

CQ: Do these programs print the same things? 2 1 A: yes B: no x = 12 if x > 10: print (x) x = x + 1 print (x) x = 12 if x > 10: print(x) else: x = x + 1 print(x)

Therefore Always ask what “equivalent” means… For us from now on: Two programs are equivalent if they give the same output for all inputs Any other meaning should be made clear by explicit wording

Some Interesting Observations Comparisons can be chained x < y <= z is equivalent to: x < y and y <= z The not operator has a lower order of precedence than comparisons not a == b is equivalent to: not( a== b)

Clicker Question Now we can start building useful conditions Does this check if x > 0? if x and y > 0: print( x, y ) A: yes B: no

Tests Continued We could write such a condition as follows: if x > 0 and y > 0: print (x + y)

Clicker Question: Are these programs equivalent? if (x+y) < 10: print(x) if (x+y)>=10: print(y) if(x+y) < 10: print(x) else: print(y) 21 A: yes B: no

Are these programs equivalent? #1 if (x+y) < 10: print(x) if (x+y)>=10: print(y) #2 if(x+y) < 10: print(x) else: print(y) (a)Yes (b)No

Nested Conditionals if x < 10: print(x+y) if y < 100: print(x) else: print(y)

Why nested conditionals? We could just write many complex conditionals and avoid the nested conditionals This is inefficient (many redundant checks) and may lead to code duplication But is the nesting always easy to understand?

Example Revisited if x < 10 and y < 100: print(x+y) print(x) if x = 100: print(x+y) print(y) if x < 10: print(x+y) if y < 100: print(x) else: print(y)

Defensive Coding Comment your code Use conditionals to check/verify assumptions Use variable names that have meaning Avoid reusing variable names Use parentheses when order matters Keep the code simple

Keeping it simple def minOfThree(a,b,c): if a<=b and a<=c: return a if b<=a and b<=c: return b if c<=a and c<=b: return c

Keeping it simple? def minOfThree(a,b,c): if a<=b: if a<=c: return a else: return c else: if b<=c: return b else: return c

Analysis def minOfThree(a,b,c): if a<=b: if a<=c: return a else: return c else: if b<=c: return b else: return c a<=b a<=c b<=c y y y n n n a b c c

Analysis a<=b a<=c b<=c y y y n n n not b! not a! a b c c Who can be min? def minOfThree(a,b,c): if a<=b: if a<=c: return a else: return c else: if b<=c: return b else: return c

Clicker Question 1 if “False”: print(“hi”) 2 if False: print(“hi”) A: 1 and 2 both print B: only 1 prints C: only 2 prints D: neither 1 nor 2 print

Clicker Question 3 if eval(“False”): print(“hi”) 2 if False: print(“hi”) A: 3 and 2 both print B: only 3 prints C: only 2 prints D: neither 3 nor 2 print

#1 if eval(“False”): print(“hi”) #2 if False: print(“hi”) (a): 3 and 2 both print (b): only 3 prints (c): only 2 prints (d): neither 3 nor 2 print

CQ explained 1: nonempty string means True 2: condition is False 3: eval(“False”) is False Therefore only 1 will print if False: print(“hi”) if eval(“False”): print(“hi”) if “False”: print(“hi”)

Error Checking Revisited #This function expects two positive integers # Returns -1 as an error condition def sumOfTwo(a,b): if type(a) == int and type(b) == int : if a > 0 and b > 0: return a+b return -1

Error Checking Revisited #This function expects two positive integers # Returns -1 as an error condition def sumOfTwo(a,b): if type(a) == int and type(b) == int : if a > 0 and b > 0: return a+b else: return -1

Two-Way Decisions import math def main(): print("This program finds the real solutions to a quadratic”) a,b,c=eval(input("\nPlease enter coefficients (a, b, c): ")) discRoot = math.sqrt(b * b - 4 * a * c) root1 = (-b + discRoot) / (2 * a) root2 = (-b - discRoot) / (2 * a) print("\nThe solutions are:", root1, root2) main()

Two-Way Decisions As per the comment, when b 2 -4ac < 0, the program crashes. This program finds the real solutions to a quadratic Please enter coefficients (a, b, c): 1,1,2 Traceback (most recent call last): File "C:\Documents and Settings\Terry\My Documents\Teaching\W04\CS 120\Textbook\code\chapter3\quadratic.py", line 21, in -toplevel- main() File "C:\Documents and Settings\Terry\My Documents\Teaching\W04\CS 120\Textbook\code\chapter3\quadratic.py", line 14, in main discRoot = math.sqrt(b * b - 4 * a * c) ValueError: math domain error

Two-Way Decisions We can check for this situation. Here’s our first attempt. # quadratic2.py # A program that computes the real roots of a quadratic equation. # Bad version using a simple if to avoid program crash import math def main(): a, b, c = eval(input("Please enter the coefficients (a, b, c): ")) discrim = b * b - 4 * a * c if discrim >= 0: discRoot = math.sqrt(discrim) root1 = (-b + discRoot) / (2 * a) root2 = (-b - discRoot) / (2 * a) print("\nThe solutions are:", root1, root2)

Two-Way Decisions We first calculate the discriminant (b 2 -4ac) and then check to make sure it is not negative. If it is, the program proceeds and we calculate the roots. Look carefully at the program. What’s wrong with it? Hint: What happens when there are no real roots?

Two-Way Decisions This program finds the real solutions to a quadratic Please enter the coefficients (a, b, c): 1,1,1 >>> This is almost worse than the version that crashes, because we don’t know what went wrong!

Two-Way Decisions We could add another if to the end: if discrim < 0: print("The equation has no real roots!" ) This works, but feels wrong. We have two decisions, with mutually exclusive outcomes (if discrim >= 0 then discrim < 0 must be false, and vice versa).

Two-Way Decisions

# quadratic3.py # A program that computes the real roots of a quadratic equation. # Illustrates use of a two-way decision import math def main(): print "This program finds the real solutions to a quadratic\n" a,b,c = eval(input("Please enter the coefficients (a, b, c): ")) discrim = b * b - 4 * a * c if discrim < 0: print("\nThe equation has no real roots!") else: discRoot = math.sqrt(b * b - 4 * a * c) root1 = (-b + discRoot) / (2 * a) root2 = (-b - discRoot) / (2 * a) print ("\nThe solutions are:", root1, root2 ) main()

Two-Way Decisions >>> This program finds the real solutions to a quadratic Please enter the coefficients (a, b, c): 1,1,2 The equation has no real roots! >>> This program finds the real solutions to a quadratic Please enter the coefficients (a, b, c): 2, 5, 2 The solutions are:

Multi-Way Decisions The newest program is great, but it still has some quirks! This program finds the real solutions to a quadratic Please enter the coefficients (a, b, c): 1,2,1 The solutions are:

Multi-Way Decisions While correct, this method might be confusing for some people. It looks like it has mistakenly printed the same number twice! Double roots occur when the discriminant is exactly 0, and then the roots are –b/2a. It looks like we need a three-way decision!

Multi-Way Decisions Check the value of discrim when 0: handle the case of two distinct roots We can do this with two if-else statements, one inside the other. Putting one compound statement inside of another is called nesting.

Multi-Way Decisions if discrim < 0: print("Equation has no real roots") else: if discrim == 0: root = -b / (2 * a) print("There is a double root at", root) else: # Do stuff for two roots

Multi-Way Decisions

Imagine if we needed to make a five-way decision using nesting. The if-else statements would be nested four levels deep! There is a construct in Python that achieves this, combining an else followed immediately by an if into a single elif.

Multi-Way Decisions if : elif : elif :... else:

Multi-Way Decisions This form sets of any number of mutually exclusive code blocks. Python evaluates each condition in turn looking for the first one that is true. If a true condition is found, the statements indented under that condition are executed, and control passes to the next statement after the entire if-elif-else. If none are true, the statements under else are performed. The else is optional. If there is no else, it is possible no indented block would be executed.

Multi-Way Decisions Conditions need not be mutually exclusive. Therefore, changing the sequence may give different results! Consider the following two versions when a==0 and b==2 if b<0: x=1 elif b>0 and a<3: x=2 elif b>0: x=3 else: x=4 if b<0: x=1 elif b>0: x=3 elif b>0 and a<3: x=2 else: x=4

Multi-Way Decisions def main(): a, b, c = eval(input("Please enter the coefficients (a, b, c): ")) discrim = b * b - 4 * a * c if discrim < 0: print("\nThe equation has no real roots!") elif discrim == 0: root = -b / (2 * a) print("\nThere is a double root at", root) else: discRoot = math.sqrt(b * b - 4 * a * c) root1 = (-b + discRoot) / (2 * a) root2 = (-b - discRoot) / (2 * a) print("\nThe solutions are:", root1, root2 )

Which version is simpler? def sort3(a,b,c): if a<=b: if b<=c: return [a,b,c] elif a<=c: return [a,c,b] else: return [c,a,b] else: if a<=c: return [b,a,c] elif b<=c: return [b,c,a] else: return [c,b,a] def sort3(a,b,c): if a<=b: if b<=c: return [a,b,c] else: if a<=c: return [a,c,b] return [c,a,b] else: if a<=c: return [b,a,c] else if b<=c: return [b,c,a] return [c,b,a]